Hi,
maybe I haven't searched enough, but I don't find anything that halps me to
solve my problem.
I've got a serverapplication. The problem is in the following lines:
application = service.Application("TwistedServerApp")
internet.TCPServer(8800, pb.ServerFactory...)
I don't want to start this with a predetermined port. More like:
twistd -y TwistedServerApp.tac 8800
Is this somehow possible? sys.argv seems impossible to use.
-Stephan